Question: $f(x) = 5x^{2}+h(x)$ $g(n) = 3n+4(f(n))$ $h(t) = 2t+3$ $ h(f(-3)) = {?} $
Answer: First, let's solve for the value of the inner function, $f(-3)$ . Then we'll know what to plug into the outer function. $f(-3) = 5(-3)^{2}+h(-3)$ To solve for the value of $f$ , we need to solve for the value of $h(-3)$ $h(-3) = (2)(-3)+3$ $h(-3) = -3$ That means $f(-3) = 5(-3)^{2}-3$ $f(-3) = 42$ Now we know that $f(-3) = 42$ . Let's solve for $h(f(-3))$ , which is $h(42)$ $h(42) = (2)(42)+3$ $h(42) = 87$